Q.Consider the non-empty set consisting of children in a family and a relation $R$ defined as $aRb$ if $a$ is brother of $b$. Then $R$ is
(A) symmetric but not transitive
(B) transitive but not symmetric
(C) neither symmetric nor transitive
(D) both symmetric and transitive

The key idea is that the relation “is a brother of” is reflexive transitive not symmetric — but here we only check symmetry and transitivity.

Step 1 – Symmetry:

If $a$ is a brother of $b$, then $b$ is a brother of $a$ only if $b$ is also male. If $b$ is female, $b$ is not a brother of $a$. Hence $R$ is not symmetric.

Step 2 – Transitivity: …

The key to this problem is to stop thinking of relations as abstract symbols and instead picture real people. The relation $R$ is defined on the set of all children in a family. That set includes both boys and girls. The statement "$aRb$" means "$a$ is the brother of $b$".

Let’s unpack what "brother" means. If $a$ is the brother of $b$, then $a$ must be male. $b$ can be male or female — a brother can have a brother or a sister. That one asymmetry is the heart of the problem.

1. Check symmetry. Symmetry would require: if $aRb$ then $bRa$. That is, if $a$ is the brother of $b$, then $b$ must be the brother of $a$. But $b$ could be a girl. A girl is not anyone's brother. So $bRa$ fails whenever $b$ is female. Even if $b$ is male, $b$ is still a brother of $a$ only if $a$ is male — which $a$ is, by definition. So for two brothers, $aRb$ and $bRa$ both hold. But the relation fails symmetry because it doesn't hold for all cases. A single counterexample is enough: take $a$ as a boy and $b$ as his sister. Then $aRb$ is true, but $bRa$ is false. Hence $R$ is not symmetric.

2. Check transitivity. Transitivity says: if $aRb$ and $bRc$, then $aRc$. Let’s test this. If $a$ is the brother of $b$, then $a$ is male. If $b$ is the brother of $c$, then $b$ is also male. So $a$ and $b$ are both boys. Now, $b$ being the brother of $c$ means $b$ and $c$ share at least one parent. Since $a$ is also a brother of $b$, $a$ and $b$ share parents. Therefore $a$ and $c$ share the same parents (through $b$). And $a$ is male. So $a$ is indeed the brother of $c$. This holds regardless of $c$'s gender. So transitivity works. …

Method: Testing a real-world relation for symmetry and transitivity

Use this for word-problem relations (brother-of, friend-of, taller-than) where you must reason about people/objects, not symbols.

Steps

Step 1: Restate the relation with the hidden attribute made explicit.

'$a$ is a brother of $b$' secretly requires $a$ to be male, with no restriction on $b$'s gender. Naming that asymmetry up front is the key to the whole problem.

Step 2: Test symmetry by looking for a case where the reverse fails.

Symmetry needs $aRb \Rightarrow bRa$ for ALL pairs. Since the family set includes girls, take $a$ a boy and $b$ his sister: $aRb$ holds but $bRa$ fails (a sister is not a brother). One such pair breaks symmetry.

Step 3: Test transitivity by following the attribute through the chain. …

Common Mistakes

Mistake 1: Calling 'brother of' symmetric.

Why it's wrong: if $a$ is the brother of $b$, then $b$ is a brother of $a$ only when $b$ is also male; if $b$ is a sister, $bRa$ fails. Correct approach: since the family has both boys and girls, a boy-and-his-sister pair breaks symmetry.

Mistake 2: Assuming the set contains only boys.

Why it's wrong: 'children in a family' includes girls, which is exactly what defeats symmetry — restricting to boys would wrongly make it symmetric. Correct approach: use the full set including sisters when testing.

Mistake 3: Treating the hidden 'male' condition as irrelevant. …